PROBABILITY
AND
MATHEMATICAL STATlSTlCS
SOJOURN TIME OF SOME REFLECTED B R O W A N
MOTION IN THE UNIT DISK
BY
* .
MADALINA D E A C O N U , MMAI GRADINARU
AND JEAN RODOLPHE ROCHE (NANCY)
Abstract. We study the heat diffusion in a domain with an obstacle inside. More precisely, we are interested in the quantity of heat
in so far as a function of the position of the heat source at time 0. This
quantity is also equal to the expectation of the sojourn time of the
Brownian motion, reflected on the boundary of a small disk contained
in the unit disk, and killcd on the unit circle. We g v e the explicit
expression of this expectation. This allows us to make some numerical
estimates and thus to illustrate the behaviour of this expectation as
a function of starting point of the Brownian motion.
1991 Mathematics Subject Classifications: 60J65, 6OH30, 35K20,
35J25, 3OC35, 30E25.
Key words and phrases: reflected Brownian motion, boundary value
problems, fractional linear transformation, numerical computations.
INTRODUCTION
Assume that a heat source is placed in a bounded domain D with a heat
absorbing boundary 8D. Also a small obstacle 0 with heat reflecting boundary 8 0
is placed inside the domain. A natural question arises: given a position of the
obstacle, what will the point z E D\O be where we must place the heat source, so
that the quantity of heat
m
Q(z):=
1 dw j dtu(t, z, w )
D\O
.m.
0
will be a maximum? It is a classical opthisation problem for linear partial
differential equations (see also [I21 and references therein). Here u (.,z , .) is the
unique solution of the heat equation with mixed boundary conditions
(au/at)(.,z,.)=(1/2)Au(.,z,.)
~ ( 0 z,
1 .) =
(aujan) (., Z , w) = 0
u(., z , w) = 0
in RTx(D\O),
for W E ~ O ,
for w E 8D.
20
M. Deaconu et
d.
The purpose of this paper is to give, using some probabilistic remarks, an
answer to this question for the simple case where D is the unit disk in the
complex plan and 0 is a small disk inside D.
To modelise heat as a probabilistic object, we need to consider the Brownian motion starting from .ZED.The reflecting and the absorbing features are
ensured by imposing that the stochastic process is reflected on 80 and killed on
8D.Then the quantity of heat is related to the sojourn time of this stochastic
process in the domain D\O. By a simple stochastic calculation we can see that
where z is the first hitting time of dD by the stochastic process. Hence, finding
the quantity of heat is equivalent to finding the preceding expectation (see also
[dl, Chapter 11).
Let us note that we may follow the potential theory point of view: the
quantity of heat is the integral on D\O of the fundamental solution (Green
-Neumann function or O-potential) associated with the stochastic process. If
D\O is an annulus centered in the origin, the Green function for the Dirichlet
problem was explicitly calculated by many authors: [13], p. 140; [3], p. 386;
[ll], p. 6.41 (see also [2], Section 11.7 and notes therein). In [6] a general
Neumann problem is aIso described (see Sections 15.2 and 15.7).At our knowledge, there is no reference for a mixed boundary problem on the annulus.
It must be noticed that, since the expression of Q(z) which we obtain is
complicated, in general, computing its maximum is not easy. We illustrate the
behaviour of this function using some numerical computations.
1. SETTING AND MAIN RESULT
1.1. S e a g . Let us consider a complex Brownian motion B starting from
a point z with lzl < 1. It is well known that the expectation of the exit time from
the unit disk of 3 is (1/2)(1, ) ' ~ z I and it is a maximum when the starting point
is z = 0.
Assume that a reflecting obstacle is placed in the unit disk. What will the
expectation of the exit time be from the unit disk of the Brownian motion
which is reflected when it hits the obstacle? Let us denote by (xp: t 2 0) the
process which is a reflected Brownian motion on an inner small circle yo, of
radius R,, killed at the first hitting time of the unit circle y,.
It is a simple calculation (see also [Ill, Chapter 6) to show that if the
circles yo and y1 are concentric, then the expectation of exit time from the unit
disk of x: is (112)(1 -]zI2 + Rg log lzJ2).This expectation is a maximum when the
starting point lies on the circle yo.
We shall also consider the process (x:: t 2 0) which is the Brownian motion reflected on the unit circle yl and killed at the first hitting time of yo (see
also Figure 1).
21
Reflected Brownian motion
Figure 1. Reflected Brownian motion on yo or on y,
We may assume, without loss of generality, that yo is centered on the real
axis. Take co E 1- 1, 1 [, Ro €10, 1 - lcol[ and denote the domain between the
circles y o and y , by
For j = 0 or 1, we shall denote by
.tj
the first hitting time of yl - by the pro-
cess .vj:
It is known that, for any
Z E 8,
for j = 0, 1, zj is finite &-as.
8.2. Main result. We are interested in computing the expectation of the
hitting time, E,(zJ, j = 0, 1, as a function of the starting point z. These functions are given in the following main result:
THEOREM
1.1. The expectations of the hitting times are given by
(1.2)
E, (z,)
= (1/2) (lz sinh p -cosh
-log
pl"lz
coshp-sinh p12)
lz cosh p - sinh pl lz q sinh 2p - rl
lz sinh p - cosh pl lz r -q sinh 2pl
and
(1.3)
E, (z,)
= (112) (lz sinh p -cash pl"
-log
lz coshp- sinh pI2)
R lz cosh p - sinh pl
lz sinh p -cosh pl lz r - q sinh 2pI2
-.
22
M. Deaconu et al.
R
lz cosh p -sinh pl lz q sinh 2p - rl
+ l o g
r2
IZ sinh p -cosh pl lz s -24 (1- q) sinh 2pl
q lz (1+ cosh 2p)- sinh 2pI2 - R4 IZ (1-cash 2p) + sinh 2pI2
r
122 r - 2q sinh 2p12
--
where we have put
. .
The sequences (sj;n(z, p, R)),3, are explicit (see (3.6)) and conuerge to zero as
R4", uniformly for z E a.
The quantities sfb are sums of logarithms and fractions of the same type as
the other terms in (1.2) and (1.3).
Let us note that for co = 0 we find the formulas of the concentric circles
case. The expectation of the exit time of the Brownian motion from the unit
disk (without obstacle) can be obtained by taking c0 = R , = 0.
1.3. Main ideas and plan. The reflected process that we study can be written, for j = 0 , 1, as
with
where nj is the outward normal at y j to 8, and, for j = 0, 1, (k!: t 2 0) are
adapted continuous and locally bounded variation processes_(see also [lo],
p. 512).
'.The expectation of the hitting time z j is then expressed as a function of the
starting point z:
Indeed, for j = 0 , 1 , let H j be a smooth function which satisfies the mixed
boundary conditions of the (Dirichlet-Neumann) problem:
23
Reflected Brownian motion
By It& formula we can write
To get (1.8) it suffices to put t = zj and to take the expectation.
Hence, it suffices to compute the function H j . But, for j = 0, 1,
(1.12)
H ~ ( z ) = ~ G ~ ~ ~ ( w ~, z€) 0~ w
, ,
P
where GjD' is-the-fundamentalsolution of the problem (1.9)-(1.11). &r j = 0, 1,
GiD) satisfiest
(1.13)
AG~Rl(w,z)=G,(z) for z ~ 0 ,
In (1.13) and (1.14) the differentiation is with respect to z.
If the circles yo and y , are concentric, the expression of this fundamental
solution can be obtained as for the reflected linear Brownian motion (see
Section 2). By using a suitable linear fractional transformation we reduce the
general case to the case of concentric circles (see Section 3.2).
The proof of Theorem 1.1 is given in Section 3.3, except for some technical
lemmas postponed to the Appendix. Finally, Section 4 presents graphs of the
expectation as a function of z.
2. CONCENTRIC CIRCLES CASE
2.1. Expectation of the hitting time. In this section we shall assume that
c, = 0,
that is, y o is centered in the origin. For the sake of completeness, we
recall some classical results (see also [Ill, Chapter 6).
PROPOSITION
2.1. The expectations of the hitting times are -&@en by
and
Note. We have already noted that (2.1) and (2.2) are nothing but (1.2) and
(1.3) with co = 0.
P r o o f of P r o p o s i t i o n 2.1. It is classical: for j
= 0,
1, the function
24
M. Deaconu et al.
satisfies (1.9) in the annulus
Then we find uj, bj such that the boundary conditions (l.lO), (1.11) are fulfilled.
Therefore (2.1) and (2.2) are obtained by (1.8). rs
It em a r k 2.2. The maximum of the function z w EE,(z,) lies on the circle
yo and the one of the function z ~ E , ( z l )lies on 7,. w
N o t e . It must be noticed that in this case we have the "intuition" of the
solution of the problem (1.9Hl.11) because the domain, the annulus A,,, is
very particular.
2.2. Fundamental solution for the anndus. We shall point out the fundamental solution of the problem (l.Bj(1.11)for the annulus ARo.This function
will be denoted, for j = 0, 1, by GIR0', and it satisfies (1.13H1.15)with A,,
instead of W.
The idea comes from the study of the linear Brownian motion on 10, I[,
reflected at 0 and absorbed at 1 (see also [4], pp, 77 and 79, or [9], p. 97). This
process has the density
~ / ~(- Ix-yI2/2t).
where q, Cy, x ) : = ( 2 ~ c t ) -exp
For the 2-dimensional case, we shall use the homothetic transformation
p(z) := z/R, and the inversion v (z) := Ri/5 instead of the translation and of the
symmetry (see also [Ill, Section 4.3). The a-potential associated with the process x i is obtained by integrating in t the product by eWatof the following
density:
where p, (w, z)
-
= (274-I
exp (- lw -zI2/2t). It is known that
.5,n.
where KO denotes the modified Bessel function of index 0. Moreover,
K,(P) log(l/P) as 8 4 0 (see, for instance, [8], p. 133).
Letting olJ0 in the expression of the or-potential for the process xj, we get
the 0-potential or the Green function
Reflected Brownian motion
I
25
To obtain the solution of (1.13H1.15) on A,,, we need to add a harmonic
function (see also [13], p. 140, 131, p. 386, Ill], p. 6.41, for the Dirichlet
problem):
PROPOSITION
2.3. For j = 0 , 1 and for w # z, we introduce
Then GjRol sati$es (1.13H1.15) on the annulus ARo.
P r o of.: We prove the result for j = 0, the case j = 1 being similar. The
function GhRo)is well defined. Indeed, the series in (2.3) is convergent. For
instance, if n E N, the series behaves as k
R p , which is convergent since
0 < R, < 1 (here k = -w/z - wZR;+ wFf wR$z).
Since, for z E yl, zZ = 1, the general term of the series in (2.3) equals 0 and
(1.15) is obvious.
For the proof of (1.14) let us put z = pis.
To compute the normal derivative it suffices to differentiate with respect to e, since the circles are centered at
the origin:
z:=,
Clearly, the series is uniformly convergent. For instance, if EN, the series
behaves as k
Rp, where k = w (1 +e2) (1 - R;)/(e2 e i 3 For Z E A,,,
Q - +
~ 1 < R, + 1, so the series converges.
Then we can verify (1.14), since, by I*),
we get
xr=,
Finally, to prove (1.13) we note that, for ~ E Zthe
, points A, of the form
r
w R p , l/wR$'+',
l/wR$", wRP+
Iie in the complementary of the annulus ARo,excepting w. Hence the functions
log ~ Z - A , / ~ and log ~ Z - A , / ~ are harmonic in A,,, and
A, G$~~O'(W,
Z) = ( 1 1 2 ~A,
) (log 12- wl +log lzl) = dw(z). H
26
M. D e a c o n u et al.
R e m a r k 2.4. The function G y O )is not symmetric in (w, 2). However, we
can prove that, for j = 0, 1,
A, GjR0)(w, Z)= 6,(w)
(2.4)
for w
EAR^.
Indeed, we note that, for ~ E Zthe
, points of the form
lie in the complementary of the annulus A,,, excepting z. Then we can proceed
as for the p r o d of (1.13). a
2.3. Find the expectation using the fundamental solution. By integrating
(2.3) on A,, we find the expectation Ez(zj),j = 0, 1, that is (21) and (2.2) up to
the multiplication by -2 (see (1.8) and (1.12)).
Let us remark first that the integrals of the terms of the series are zero
except for the term corresponding to n = 0. Indeed, this is a consequence of the
facts that, for j = 0, 1 ,
Iz/RpI,
I~/zR~'"-"~
1, l l / ~ R ~ - ~IZ/RF+~I
jl,
are
{>
1
<Ro
if n ~ z $ ,
ifn~Z*_,
and of the following
LEMMA2.5. We have
J
(2.5)
log][ -11' d[
A ~ o
=
2n (1-Ri) log IR]
tf I4 > 1,
- 2x Rg log 1
1
1-n: (1 - lRl2)
-2n:R$logRo-~(1-R~)
if Ro < 1
1
1< 1,
ifljll<Ro.
The proof of this lemma is postponed to the Appendix.
On the other hand,
Since, for j = 0, Ro < lzl < 1, Izl/Ri > 1, 1/15]R i > 1 and 1/IZI > 1, using again
(2.5) we get
1 GbRO)(w, Z)dw = (1124 log ]zl j dw +(1/4n) ( -271 Ri log 121- .rc (1- 1zI2)
ARO
+ 2n (1 -Ri) log (l/IL;IR@-2n (1- Ri) log (l/IZI) -2n (1- R t ) log (IZI/R;))
ARO
(1~1'-
1-2 log 121) = (- 1/2) EZ(70).
= (1/2)(1- ~ i ) l o glzl+ (114)
The calculation is similar for j
=
1, by noting that
~ $ 1 ~ . 1 < Ro. H
27
Reflected Brownian motion
3. FRACTIONAL LINEAR TRANSFORMATION AND GENERAL CASE
3.1. Fractional h e a r transformation. Let us consider, for t E R , the transformation
[ cosh t sinh t
[€ C .
'":=
(sinh t+cosh t'
+
This is a one-parameter transformation group: a,a, = a,,,. We recall below
some well-known properties of these transformations.
R e m a r k 31. The image of the circle centered in the origin,-with radius
tanh r, by the fractional linear transformation a, is a circle centered on the real
axis. Moreover, this transformation leaves the unit circle invariant.
Indeed, it is no difficult to see that lat(l;)l = 1, provided lil = 1. On the
other hand, by classical properties of the fractional linear transformations, the
image of an orthogonal circle to the real axis is a circle orthogonal to the real
axis. Therefore, the center of the image circle lies on the real axis.
Moreover, the images of the points tanh r and -tanh r are tanh (t r)
and tanh(t -r), respectively. Hence the image circle has the center
(1/2) (tanh (t $ r ) tanh (t - r)) and the radius 11/21( t a d (t r) -tanh (t - r)).
In particular, the circle yo, centered in cO, with radius R,, is the image of
a circle centered in 0 having the radius R, by the transformation of parameter p,
where R and p are given by (1.4). rn
+
+
+
R e m a r k 3.2. Let a and a-I be the solutions of the equation
By a geometric reasoning we can see that, for j
(3.2)
l(i-a)/([-a-l)l=const
= 0,
1,
provided ( € y j .
Therefore, yo and y , are in the family of Apollonius circles with respect to oc and
a-l (see [I], p. 84). There exists an orthogonal family of circles which pass
through the points a and a-l, and satisfy
-
(3.2')
.
-
(see also
arg ((-a)/((
-
[n, p. 53).
-a - I ) = const
7
3qp
ra
3.2. Fundamental solution for the general domain. We shall point out the
fundamental solution of the problem (1.9)-(1.11) using the one on the annulus
A,, and the linear transformation defined above (see also [Il], p. 6.19):
PROPOSI~ON
3.3. For j = 0, 1 and for w # z we introduce
where G$,) is given by (2.3) with the parameter R giuen in (1.4). Then G)*)satiSJies
(1.13H1.15).
28
M. Deaconu et al.
Proof. Let us first notice that
(see also [ll], p. 4.29). It follows, by a reasoning similar to that for the proof of
Proposition 2.3, that
d=G$al(w, z) = A, G$l(a-,(w), a-,(z))
= (1/2n)Az(logla-,(z)-
(w)l+log (la-, (z)l/Rq).
Since 4 is a holomorphic function, the second term in the above sum is a harmonic function. Then, by (3.11, we get
a.-,
Therefore,
A, Gj(w, z) = (1/2n) A , (log lz -wl -log I(l/tanh p) -zl)
and we get (1.13), since tanh p < 1.
In order to obtain (1.14) we use the corresponding property of GjR),(*) and
Remark 3.2.
Indeed, to calculate the derivative of GjQ) in the normal direction to the
circle y j we can use an infinitesimal shift along an arc of circle from the orthogonal family given by (3.2'). We have already observed that a _ , send the
Apollonius circles (3.2) in the concentric circles in 0 and the orthogonal family
(3.2') in straight lines through 0 (that is, into two other orthogonal families of
circles in wider sense; see [I], p. 79). Therefore, using an infinitesimal shift
along a segment on a line through 0, the upper derivative is equal to the
derivative of GjR)in the normal direction to the image circle centered in 0. But
this derivative is zero and (1.14) follows.
Finally, (1.15) is a consequence of (*) and the corresponding property
of GY).
3.3. Expectation of the hitting time: the proof of Theorem 1.1. As in Section 2.3 we shall find the expectation of the hitting time using-(1.8), (1.12) and
.
the fundamental solution obtained in Proposition 3.3.
By (3.3) and (*) from the proof of Proposition 3.3, we can write, for
j = 0 , 1,
m-
where Jac(r) is the Jacobian of the transformation a,. Let us write this Jacobian as
29
Reflected Brownian motion
where k
=
- l/tanh p < - 1. Therefore,
Using Green's formula, we obtain
where the normal derivatives and the surface integral are with respect to I .
By (2.41, the first integral in (i) is the same for j = 0, 1, and equaIs
(ii)
A,
=
1
4sinh4p la-,(z)-k12
1
1
tanh2 p
- 4 s i n l 1 ~ Il+
~ a-,(z)tanhp12'
-
By (2.31, the second integral in (i) can be written as
where
t'"
Since the derivatives are in [, the first integral in (iii) equals
A#
=
a 1
J ---1 6 sinh4p
~
,,,,=,,an 15-kI2 dc- (lil=s
an 15-kI2
(I--
la-, (z)I2
To compute the preceding quantity we shall use the following
LEMMA3.4. For 0 < e 4 1 ,
30
M. Deaconu et al.
We postpone the proof of this lemma to the Appendix, and we use its result to
obtain
since k2- 1 = l/sinh2p and here r is given in (1.5).
The second term in (iiij can be written as
Ay'
(v)
-
1
m
1 6 7 ~ s i n hC
~ ~ {(J
~ = - ,,
(z)/R4n)- ( R a - (z)/R~~))
where
The expression of J ( Q , 1) is contained in the following result, the proof of
which will be given in the Appendix:
LEMMA3.5. For 0 < q
< 1,
Recalling that l / k = -tanhp, by (vi) and (3.5) we can compute the genera1
term of the series in (v):
(vii)
-
-.
J(1, A)-J(R, A)
16n sinh4p
R
$log IA+ tanhpl --log
13,+R2tanhpl if IAI > 1,
2r2
R2
1
1- Ill2tanh2 p
$log11 +Atanhpl-Z-iloglA+R2 tanhptr
4 sinh2p I1 + A tanh p12
31
Reflected Brownian motion
Here q and r are given in (1.5), and
A E ( ~ - ~ ( Z ) / R l~/"i ,i - , ( ~ ) R ~ ~ ,+ 1~/G-p(~)R
-~j
4"- 2 j, a-,(z)/R ~ J I + Z .
n ~ Z , j =0 , 1).
In order to apply (vii) for the calcuIation of A$%,, we need to study the
modulus of the complex in the preceding set for n € Z and for j = 0, 1.
For n = 0,
Therefore,
(viii)
1
Ayj, = log
11+a-,(z) RZtanhpl 1
1- ] a - , ( ~ ) 1 ~tanh2p
la-, (z) R2 tanhpl 4 sinh2pl 11 +al, (z)tanhpI2
+
RZ
2rZ
--- log
+
la-, (z) R2 tanhpl 11 +a_, (z) R4 tanhpt
I1 +a-,(z) R2tanhpl la-,(z)+R4 tanhpl
and
(ix)
1
= log
R2 11+a-,(z)tanhp12
la-,(z)+RZtanhp12
-
1
1--la_,(~)1~
tanh2p
4sinh2p Il+a-,(z)tanhp12
RJl+a-,(z)R2tanhpl
la-,(z)+R4 tanhpl
R'
--log
2r2
For ~ E Z ' ;and j
= 0,
+-.2rq
la-,(~)1'-R~tanh~~
la-,(z)+R2 tanhp12 '
1,
la-, (z)/R4"l, I l / ~ - ~ (R4n+2-2j
z)
1, ll/CT-p(~)R4n-2il,l a - , ( ~ ) / R ~ " +>
~ 11,
while, for n ~ZT and j
= 0,
1,
, j = 0, 1, the computations give
Thus, for ~ E Z *and
~<-,
ABn+A"(-~ =
+log
(a-,(z)+ R4"tanhpl 11+a-,(z) R4"+Z-2jtanhpl
log 11+a-p (z)~4n-2jtanhpl
+ ~ 4 " + tanhpl
2
I1 +a-,(z) R4"tanhpl ( a - , ( z ) + ~ ~ " - ~ + ~ j t
la-,(z)+R4"+ 2jtanhplIl +a-,(z) R4n-2tanhpl
32
M. Deaconu et al.
where q and r are given in (1.51, and a_,(z) = (z coshp-sinhp)/( -z sinhp
coshp).
Combini,ng (i), (ii), (iv), (vii) and (x) we get
+
while from (i), (ii), (iv), (ix) and (3.6) we get
m
(xii)
H I M = AI+.~$~?+A$YO+ ( A L ~ ~ ~ + A ~ ~ ~ ( - ~ J .
n=l
Thus (1.2H1.3)are obtained by using (1.8) and (x), where we have put
In order to end the proof, let us show that, for j = 0, 1, the series with
general term given by (3.6) are convergent. For instance,
has the same behaviour as k
znm=
R4", where
=,
The other series with logarithm can be treated in the same manner. Similarly,
-
1 - l a - , ( ~ ) 1 ~ RtanhZp
~~
l a - , ( ~ ) 1 ~ - R ~ " -4jta
~ +nh2p
[l+a-,(z) R4"tanhp12 la-,@)+ R4"-2+2jtanhp12
+
( z~)2~R E R - 4tanhp
- l a - , ( ~ ) 1 ~ - R ~ ~ + ~-~l -t ~a an-hp ~
I a - , ( z ) + ~ ~2jtanhp12
~+
11+a-,(z) R4n-2tanhp12
behaves as k
zn",,R4", where
k
= (-2
tanhp)(l -R-2)(la-,(z)l-
R2j/la-,(z)l).
This ends the proof of the theorem except for the proofs of the lemmas.
ra
Reflected Brownian motion
33
4. NUMERICAL RESULTS
In this section we present numerical approximations of the analytical
representation of the expectations given in (1.2) or (1.3). We plot two views of
the expectation as a function of the starting point: three-dimensional graph and
its vertical section.
Figures 2 and 3 illustrate the case j = 0. Recall that for the concentric
circles case the maximum lies on yo. Suppose now that co # 0. When Ro is
small, the position of the maximum is close to zero as we can see in Figure 2.
The increase df -Ro gives a displacement of this position towards- yl. This is
quite differedtwith respect to a deterministic motion in which it is quite natural
to think that z must be far from this circle.
Figures 5 and 6 correspond to the case j = 1. The maximum lies on y , and
its value is a decreasing function with respect to Ro.
In parallel to the semi-analytical method present in this paper we consider a classical finite element method to solve the partial differential equations
(1.9), (1.10) and (1.11).A solution of this problem is in H2(a),the SoboIev space
of function u such that td, VU and Au belong to LZ(i2). By the finite element
we compute an approximation of H j belonging to a subspace r/h of continuous
functions piecewise linear in Q. The results obtained by the finite element
method are similar to those obtained by the semi-analytical method considered in this paper. One difference between the two methods is that the maximum
obtained by the finite element method is an approximation with an error
of order h2, h being the discretisation parameter. In Figures 4 and 7 we represent results computed by the finite element technique, which correspond
to Figures 2b and 6b, respectively.
Figure 2. Case j = 0 with co = 0.1 and R o = 0.01
3 - PAMS 20.1
34
M. D e a c o n u et aI.
Figure 3. Case j = 0 with c, = 0.2 and Ro = 0.4
Figure 4. Case j = 0 with co = 0.1 and Ro = 0.01, by the finite element technique
Figure 5. Case j
=
1 with c,
= 0.3 'and
R,
= 0.01
35
Reflected Brownian motion
Figure 6. Case j = 1 with c,
= 0.3
and R , = 0.4
Figure 7. Case j = 1 with co = 0.3 and Ro = 0.4, by the finite element technique
APPENDIX
We give here the proofs of lemmas which we used.
Proof of Lemma 2.5. Let us put [:= eeie, A : = ae" and, for a # 0,
a := (qja)sinz and b :=(@/a) cos z. We can write
-
Since
2a
J d e log (1 +a2 + b2-2a cos 8 -2bsin8)
0
(see also [ 5 ] , p. 528, 4.225.4), we obtain
= 2nlI,2 +,23
,)
log (a2+b2)
36
M. Deaconu et at.
2~
(ltede) logo2
if c r > I,
~lc(P,~~d~)lo~~'+2n~
if ~Rol o< g e<~1 ,~ e
2n
if
@log Q d@
I T <R,.
The preceding equality
lies also for a = 0, and we get (2.5).
..
- . .
Prqof of Lemma 3.4. Let us put [:= pie
and a:= ~ / k< 1. We can write
From this we get (3.4). H
For the proof of Lemma 3.5 we need the following result:
LEMMA
A.1. FOPREC,
2 x ~ log IR,-g2/kI2
if 121 > Q,
k2-Q' {loge211-A/kl/k2 ijIAI<g.
Proof. Let us put [:=eeie, A:=aeiT, a : = ~ / k < l and, for a # O ,
b: = g/(r. We can write
da
(A.1)
=
a. 2= do
logl~-~12 = log (r2
( ~ . g = ~15-kI2
l
k o 1+a~-2acose
r:= J
The first integral is equal to (2,rclog02)/(1-a2). Then (A.l) is easily obtained for
real A (that is, for z = 0 or z = x) since
<,n-
7
(see also [ 5 ] , p. 594, 4.397.16).
Then, by a classical argument of analytic continuation, we obtain
The preceding equality lies also for
but (A.1). EJ
(r
=1
1
1= 0
and its real part is nothing
37
Reflected Brownian motion
P r o o f of Lemm a 3.5. We shall use the same notation as in the proof of
Lemma A.1. We write J(Q, A) = 11-I,, where
2x
1 d0 log (Q' +02-2ea
=p
cos0 cosr - 2 ~ asin8 sinr)-
o
a 2"
1d0 log
(e2
kZo
..
1
"Q
k2 -2ke cos 8
+ g2- 2 ~ acos 0 cosz - 2 ~ asin8 sin^) (1 +-2a+2cosO
a2- 2a cos 812
=-
1
12:=
d
(161
i
--
..
a
II -k12 an
log )(-a12 dcT
+
L/
-log (e2 n2 - 2 p cos8 cos z -2~r3sin8 sinz)
d0e2+k2-'2kQcosOap
=Q
2%
=bjdQ
1
2b-2cosBcosr-2sinesinz
e2+ k2- 2ke cos 8 1+ b2-2b cos 8 cosz- 2b sin0 sin^
But (A.l) can be written as
2x
(A.2)
Id0
o
log (e2+ crZ -2ea cos 0 cosz -2pcr sin0 sinz)
1+a2-2acos0
log e2(l/b2+ a2-2 (a/b)cosz)
--. 2n
I-a2 {loge (1+a2/b2-2 (a/b)cosz)
if 13Ll > Q,
if IRI < Q
or as
By the derivation of (A.2) and (A.3) with respect to a and b, respectively, we get
4ne2
e2- lAlk cos z 8xe2,,logll-e2/kl
k2(k2-e2)' I
l -e2/k12
(k2-Q )
4n
Ihl (In1-k cos 2)
8ne2
,logeIl-~/kl
k2(k2-e2)
If-)LlkI2
(k2-el
+
+
,
ifIRI>e,
if
14 < e,
38
R1. D e a c o n u et al.
and
47q2 e2- llllk cos .t
k2(k2-Q2)' Il-g2/k12
-. 4~c 1 - IL/klcos z
k2 -e 2 11 -A/k12
and the proof of Lemma 3.5 is done. ra
if 1
1
1> e,
if I4 < a,
Acknowledgements. The authors are grateful to Bernard Roynette for
stimulating discussions and helpful comments, and to Jean-Philippe Anker and
Pierre Vallois for their suggestions.
I:
REFERENCES
[I] L. Ahlfors, Complex Analysis, McGraw-Hill Book Company, New York 1966.
[2] R B. Burckel, An Introduction to Classical Complex Analysis. I , Academic Press, New
York-San. Francisco 1979.
[3] R. C o u r a n t and D. Hil ber t, Methods of Mathematical Physics. I, Interscience Publishers
Inc., New York 1953.
141 E. B. D y n k i n and A. A. Y u s hkevich, Markov Processes, Theorems and Problems, Plenum
Press, New York 1969.
[5] I. S. G r a d s h t e y n and I. M. R yzhik, Table of Integrals, Series and Products, Academic
Press, New York 1980.
[6] P. H e n r i ci, Applied and Computational Complex Analysis. III, Wiley, New York-London
1986.
[7] E. Hille, Analytic Function Theory. I, Ginn and Company, Boston 1959.
[8] K.I t 8 and H. P. McKean Jr., DifSuon Processes and Their Sample Paths, Springer, Berlin-Heidelberg-New York 1974.
[9] I. K a r a t z a s and S. E. Shreve, Brownian Motion and Stochastic Calculus, Springer, Berlin-Heidelberg-New York 1991.
[lo] P. L.L i o n s and A. S. Szni tman, Stochastic Diflerential Equations with Reflecting Boundary
Conditions, Comm. Pure Appl. Math. 37 (1984), pp. 511-537.
[11] M. Rao, Brownian Motion and Classical Potential Theory, Lecture Notes Ser., No. 47, Matematik Institut of Aarhus Universitet, 1977.
[I21 J. R. R o c h e and J. So k o l o w s k i, Numerical methods for shape ident$cn_tion problems, Control Cybernet. 25 (1996), pp. 867-894.
[I31 H. Villat, Le probldme de Dirichlet dans une aire annulaire. Rend. ~ i r c . " ~ aPalermo
t.
33
(19121, pp. 134-175.
Institut de Mathkmatiques Elie Cartan
Universiti Henri Poincark
B.P. 239, 54506 Vandoeuvre-16s Nancy Cedex, France
e-mail: [email protected]
Received on 14.7.1999;
revised version on 7.2.2000